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Figure from:Didaktik der Physik-Beiträge zur DPG-Frühjahrstagung-Jahresband 2020

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Figure 16: Playing with the lenses. Upper: (left) our fac- ulty building through the Plummer sphere lens, (right) a Figure 15: The Hubble Ultra Deep Field (Credit:ESA/ Hubble NASA) seen through the point mass lens on the left. The lensed group of galaxies "Cheshire Cat" on the right. Credit for the right image: NASA/STSclI.

Figure 16 Playing with the lenses. Upper: (left) our fac- ulty building through the Plummer sphere lens, (right) a Figure 15: The Hubble Ultra Deep Field (Credit:ESA/ Hubble NASA) seen through the point mass lens on the left. The lensed group of galaxies "Cheshire Cat" on the right. Credit for the right image: NASA/STSclI.

Related Figures (13)

Figure 1: Representation of the gravitational-lens effect (not to scale) of the system consisting of a distant star, the Sun and the Earth. We can see the indicated deflection angle 5 . Credit: path2space.com, elaborated by S. Simionato. First of all, for simplicity and to reach our educa- tional goals without distracting the attention of our students with intricate calculations, all gravitational lenses we consider in our models are axially sym- metric.
Figure 1: Representation of the gravitational-lens effect (not to scale) of the system consisting of a distant star, the Sun and the Earth. We can see the indicated deflection angle 5 . Credit: path2space.com, elaborated by S. Simionato. First of all, for simplicity and to reach our educa- tional goals without distracting the attention of our students with intricate calculations, all gravitational lenses we consider in our models are axially sym- metric.
Figure 2: Geometry of the general situation of the gravita- tional-lens effect. The line connecting observer O and lens L is the starting point for defining the angles that corres- pond to the position of the source 0, and the position of the image I, on the upper part of the diagram 0, . The second image I, is on the opposite side of this line. 6 is the angle of deflection and b the impact parameter in the lens plane.
Figure 2: Geometry of the general situation of the gravita- tional-lens effect. The line connecting observer O and lens L is the starting point for defining the angles that corres- pond to the position of the source 0, and the position of the image I, on the upper part of the diagram 0, . The second image I, is on the opposite side of this line. 6 is the angle of deflection and b the impact parameter in the lens plane.
Figure 3: Graphical Geogebra solution of eq.(5) for the point mass lens in the special case of 8,=0 . The ob- server sees a ring of radius 0, , the famous Einstein ring, which is shown in the bottom panel. Solving eq.(5), we obtain the number and position of produced images from a background point mass source. We can solve this quadratic equation analyt- ically, becoming a nice maths practice, but the easi- est way is to solve it graphically, as we did in fig. 3 thanks to the software Geogebra, and analyse then different relative positions of the source.
Figure 3: Graphical Geogebra solution of eq.(5) for the point mass lens in the special case of 8,=0 . The ob- server sees a ring of radius 0, , the famous Einstein ring, which is shown in the bottom panel. Solving eq.(5), we obtain the number and position of produced images from a background point mass source. We can solve this quadratic equation analyt- ically, becoming a nice maths practice, but the easi- est way is to solve it graphically, as we did in fig. 3 thanks to the software Geogebra, and analyse then different relative positions of the source.
The next stage in this process is to consider different types of source, for instance an extended source, and to solve the lens equation for each point of the source in order to obtain the images produced by the lens. We did this using Geogebra for an extended disk source and also for a line source. Figure 5: Simulation with Geogebra of the solution of the lens equation for the point mass lens in the case of an ex- tended disk and a line as sources, indicated with S. The se- quence starts with the upper left panel, where 0,=0 , then going right and bottom left, then right again we see the sources S moving sideways away from the lens L. The dots represent the produced images of specific points of the sources and give us an idea of the total images shape.
The next stage in this process is to consider different types of source, for instance an extended source, and to solve the lens equation for each point of the source in order to obtain the images produced by the lens. We did this using Geogebra for an extended disk source and also for a line source. Figure 5: Simulation with Geogebra of the solution of the lens equation for the point mass lens in the case of an ex- tended disk and a line as sources, indicated with S. The se- quence starts with the upper left panel, where 0,=0 , then going right and bottom left, then right again we see the sources S moving sideways away from the lens L. The dots represent the produced images of specific points of the sources and give us an idea of the total images shape.
Figure 4: Graphical Geogebra solutions of eq.(5) for the point mass lens in case of 0,>0 . The observer sees two images, one inside and one outside the Einstein radius 0, , as shown in the bottom panel. The further away the source is from the lens, as seen in projection, the closer the image will be to the source itself and the other to the lens, the latter two arrive in extreme cases to merge so that they cannot be distinguished. for reasons of full symmetry we see a ring of radius 0, . Then we continue with 0,>0 in fig. 4, where the source is considered gradually further and further away from the line joining observer and lens and where we will always see two images of the source.
Figure 4: Graphical Geogebra solutions of eq.(5) for the point mass lens in case of 0,>0 . The observer sees two images, one inside and one outside the Einstein radius 0, , as shown in the bottom panel. The further away the source is from the lens, as seen in projection, the closer the image will be to the source itself and the other to the lens, the latter two arrive in extreme cases to merge so that they cannot be distinguished. for reasons of full symmetry we see a ring of radius 0, . Then we continue with 0,>0 in fig. 4, where the source is considered gradually further and further away from the line joining observer and lens and where we will always see two images of the source.
Figure 7: Graphical Geogebra solutions of eq.(8) for the Plummer sphere lens in case of 0,>0 . The observer sees at first three images, two inside and one outside the Ein- stein radius 0, , as shown in the left bottom panel. The further away the source is, as seen in projection, the closer one image will be to the source itself. The other two in- stead move towards each other, until they merge together into a single image (central panel), then disappear (right panel). In fig. 6, the perfect observer-lens-source alignment is shown, where we not only see a ring of radius 0, , but also a central point-like image. Then we con- tinue with 0,>0 in fig. 7, where the source is con- sidered gradually further and further away from the line joining observer and lens. This time, depending on how far away the source is from the lens, we will see respectively three, two or one image of the source.
Figure 7: Graphical Geogebra solutions of eq.(8) for the Plummer sphere lens in case of 0,>0 . The observer sees at first three images, two inside and one outside the Ein- stein radius 0, , as shown in the left bottom panel. The further away the source is, as seen in projection, the closer one image will be to the source itself. The other two in- stead move towards each other, until they merge together into a single image (central panel), then disappear (right panel). In fig. 6, the perfect observer-lens-source alignment is shown, where we not only see a ring of radius 0, , but also a central point-like image. Then we con- tinue with 0,>0 in fig. 7, where the source is con- sidered gradually further and further away from the line joining observer and lens. This time, depending on how far away the source is from the lens, we will see respectively three, two or one image of the source.
Figure 6: Graphical Geogebra solution of eq.(8) for the Plummer sphere lens in the special case of 0,=0 . In this case the observer sees the Einstein ring of radius 0, and a central image, as shown in the bottom panel. Also in this case, solving eq.(8) we obtain the num- ber and position of produced images from a back- ground point mass source. The analytical solution involves a cubic equation, which engages students in a nice calculation, but again the easiest solution is graphical (fig. 6).
Figure 6: Graphical Geogebra solution of eq.(8) for the Plummer sphere lens in the special case of 0,=0 . In this case the observer sees the Einstein ring of radius 0, and a central image, as shown in the bottom panel. Also in this case, solving eq.(8) we obtain the num- ber and position of produced images from a back- ground point mass source. The analytical solution involves a cubic equation, which engages students in a nice calculation, but again the easiest solution is graphical (fig. 6).
We can of course apply this procedure to different kinds of sources, as done before, and some results for the extended disk source are shown in fig. 8. Figure 8: Simulation with Geogebra of the solution of the lens equation for the Plummer sphere lens in the case of an extended disk S as source. We can follow the sequence al- ways from left to right and going then down. The se- quence starts with the upper left panel, where 6,=0 , then we see the source S moving sideways away from the lens L. The dots represent the produced images of specific points of the source and give us an idea of the total images shape. The images of the source central point along the @-axis are always indicated with 0, , 0, and 0, .
We can of course apply this procedure to different kinds of sources, as done before, and some results for the extended disk source are shown in fig. 8. Figure 8: Simulation with Geogebra of the solution of the lens equation for the Plummer sphere lens in the case of an extended disk S as source. We can follow the sequence al- ways from left to right and going then down. The se- quence starts with the upper left panel, where 6,=0 , then we see the source S moving sideways away from the lens L. The dots represent the produced images of specific points of the source and give us an idea of the total images shape. The images of the source central point along the @-axis are always indicated with 0, , 0, and 0, .
Figure 9: Geometry of the deflection of a light ray travel- ling through two mediums with different refraction index n, n,=1 is air, according to Snell's law and introducing the concept of deflection angle 6 . tion? For answering this question we need to com- bine what we know from general relativity and op- tics. This procedure is not so complex as it seems. In fact, from experience, it is feasible to perform it with students of the last years of secondary school, espe- cially if they are particularly interested in the sub- ject. Thus, referring to fig. 9, we apply now this method and extrapolate the profile for the plexiglass enses representing a point mass and a Plummer sphere distribution of mass. These lenses will repro- duce the images we saw in the first part of this art- icle. First of all, one side of our plexiglass lens is flat for simplicity and we have to infer the shape for the other side. Secondly, our lenses are rotationally sym- metric, therefore it is enough to calculate the profile in two dimensions and then rotating it around a cent- ral axis we get the entire lens. Indeed, this simplifies our work considerably. The inclined black line in fig. 9 represents the tangent line to the surface of the plexiglass lens in the point P, if we can understand how the tangent line changes for all points of the lens, it is possible to find the corresponding function which describes the shape of the lens. Basically, what we need is the deflection angle 6 from Gen- eral Relativity and Snell's law from optics. Let's start analysing fig. 9 and introducing these concepts. We know that, when passing near a massive object, the path of a light ray from a distant source is deflected. This path is represented by the red arrows in the dia- gram. The amount of deflection, as we know and as it is shown in the figure, is represented by the deflec- tion angle 4 expressed by eq.(1). If we now apply Snell's law to this configuration and make use of the trigonometric angle sum and difference identities,* we obtain
Figure 9: Geometry of the deflection of a light ray travel- ling through two mediums with different refraction index n, n,=1 is air, according to Snell's law and introducing the concept of deflection angle 6 . tion? For answering this question we need to com- bine what we know from general relativity and op- tics. This procedure is not so complex as it seems. In fact, from experience, it is feasible to perform it with students of the last years of secondary school, espe- cially if they are particularly interested in the sub- ject. Thus, referring to fig. 9, we apply now this method and extrapolate the profile for the plexiglass enses representing a point mass and a Plummer sphere distribution of mass. These lenses will repro- duce the images we saw in the first part of this art- icle. First of all, one side of our plexiglass lens is flat for simplicity and we have to infer the shape for the other side. Secondly, our lenses are rotationally sym- metric, therefore it is enough to calculate the profile in two dimensions and then rotating it around a cent- ral axis we get the entire lens. Indeed, this simplifies our work considerably. The inclined black line in fig. 9 represents the tangent line to the surface of the plexiglass lens in the point P, if we can understand how the tangent line changes for all points of the lens, it is possible to find the corresponding function which describes the shape of the lens. Basically, what we need is the deflection angle 6 from Gen- eral Relativity and Snell's law from optics. Let's start analysing fig. 9 and introducing these concepts. We know that, when passing near a massive object, the path of a light ray from a distant source is deflected. This path is represented by the red arrows in the dia- gram. The amount of deflection, as we know and as it is shown in the figure, is represented by the deflec- tion angle 4 expressed by eq.(1). If we now apply Snell's law to this configuration and make use of the trigonometric angle sum and difference identities,* we obtain
where 6 is expressed by y instead of 9 with y=60D, and the small-angle approximation is used. At this point, solving this differential equation with the desired mass distribution inserted in the place of M(y) gives the profile of the corresponding lens. Indeed, for mass distributions not too complic- ated, this differential equation is easily solved separ- ating the variables and integrating.
where 6 is expressed by y instead of 9 with y=60D, and the small-angle approximation is used. At this point, solving this differential equation with the desired mass distribution inserted in the place of M(y) gives the profile of the corresponding lens. Indeed, for mass distributions not too complic- ated, this differential equation is easily solved separ- ating the variables and integrating.
Figure 13: Example of Einstein rings. In the centre the lensed galaxy "Cosmic Horseshoe", on the left side the Plummer sphere lens in action with a disk source and on the right side the point mass lens in action with a point source. Credit for the central image: ESA/Hubble NASA. lenses and engage students without making them lose their attention. Moreover, it is also interesting to compare the gravitational lens effects that we pro- duce thanks to the lenses with real ones or even with the Geogebra simulations. Few examples are in fig- ures 13, 14, 15 and 16.
Figure 13: Example of Einstein rings. In the centre the lensed galaxy "Cosmic Horseshoe", on the left side the Plummer sphere lens in action with a disk source and on the right side the point mass lens in action with a point source. Credit for the central image: ESA/Hubble NASA. lenses and engage students without making them lose their attention. Moreover, it is also interesting to compare the gravitational lens effects that we pro- duce thanks to the lenses with real ones or even with the Geogebra simulations. Few examples are in fig- ures 13, 14, 15 and 16.
Figure 14: Example of arcs. In the centre the lensed galaxy SDP.81, on the left side the point mass lens in ac- tion with a disk source and on the right side a Geogebra simulation with a disk source. Credit for the central image: ALMA.
Figure 14: Example of arcs. In the centre the lensed galaxy SDP.81, on the left side the point mass lens in ac- tion with a disk source and on the right side a Geogebra simulation with a disk source. Credit for the central image: ALMA.