# An angled photon (special relativity)

• philnow
In summary, the photon's velocity in S' is calculated to be (Ccos(theta) + V) / (1 + (Ccos(theta)*V/C^2)) and the Y component of the photon's velocity in S is (C*sin / (1 + cv*cos/c^2)).
philnow

## Homework Statement

A photon moves at an angle theta with respect to the x' axis in the frame S'. Frame S' moves with speed v with respect to frame S (along the x' axis). Calculate the components of the photon's velocity in S and verify that it's speed is c.

## The Attempt at a Solution

I break down the photon speed in S' into Ccos(theta) for X and Csin(theta) for y. There is no change from the transformation of frames with respect to the y-axis, only the x-axis. So the speed of the horizontal component of the photon in S frame is then

(Ccos(theta) + V) / (1 + (Ccos(theta)*V/C^2))

I would like to show that this squared, plus Csin(theta) squared should be equal to C^2. The problem is the algebra... whatever I do, I end up with cosines to odd and even powers that just won't simplify, so here I am. Any hints?

well, this part is right

$$X' : \ c_x= c\cos\theta$$
$$Y' : \ c_y= c\sin\theta$$

but there is a change in both vector components when you transform from $$S'$$ to $$S$$
(there is no change from the transformation of frames with respect to the y-axis only when $$\theta=0 \ or \ \pi$$)

as for what you're trying to do, obviously:

$$(\frac{c\cos\theta + v}{1+\frac{vc\cos\theta}{c^2}})^2 + (c\sin\theta)^2 \neq c^2 \ (i)$$

because

$$(\frac{c\cos\theta + v}{1+\frac{vc\cos\theta}{c^2}})^2 + (c\sin\theta)^2=\frac{c^2\cos^2\theta +v^2+2vc\cos\theta+c^2\sin^2\theta+2vc\cos\theta\sin^2\theta+v^2\cos^2\theta\sin^2\theta}{\frac{c^2+2vc\cos\theta+v^2\cos^2\theta}{c^2}}=$$

$$=c^2\frac{c^2+v^2+2vc\cos\theta(1+\sin^2\theta)+v^2\cos^2\theta\sin^2\theta}{c^2+2vc\cos\theta+v^2\cos^2\theta}$$

$${not} (i) \ \Leftrightarrow \ (v=0) \vee (\cos\theta=1\Leftrightarrow \theta\in \{0,\pi \})$$

which means that either $$S'$$ is not moving relative to $$S$$ either the velocity vector of the photon is parallel with the x-axis (and both cases are contradictory with the premises of the exercise).

you must go back to square one. you probably forgot how to calculate the photon's velocity vector components in S. you should have that in your textbook. good luck.

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Ok that makes sense. But how does the Y component of the photon's velocity change from S' to S?

philnow said:
Ok that makes sense. But how does the Y component of the photon's velocity change from S' to S?

well , (you should have this in your textbook but...) the Y component from S is:

$$\displaystyle \frac{c_y}{\lambda (1+\frac{vc_x}{c^2})}$$, where $$\lambda$$ is the Lorentz factor...

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Hi again. I'm assuming you meant without the lorentz factor, I think the that is only for S to S', and not vice versa. So using this, I'm getting

Ux = C*sin / (1 + cv*cos/c^2)

and Uy = C*cos / (1 + cv*cos/c^2)

Once again, my algebra defeats me. This time it simplifies much better, though... I'm getting:

C^2*(c^2 + v^2 + 2vc*cos)/(c^2 + cos^2*v^2 + 2vc*cos)

I can't get rid of that damn cos^2 in front of the v in the denominator. I realize this is trivial algebra, but I can't figure out the problem! I checked with my professor and we have the same Ux and Uy.

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Ux^2 + Uy^2... should equal C^2

What's the problem folks :(

## 1. What is an angled photon in special relativity?

An angled photon in special relativity is a photon (a type of particle that carries electromagnetic energy) that moves at an angle relative to an observer's frame of reference. This angle is due to the effects of special relativity, which describes how motion and time are perceived differently by observers in different frames of reference.

## 2. How does special relativity affect the behavior of angled photons?

Special relativity predicts that angled photons will exhibit time dilation and length contraction, meaning that they will experience time and space differently than observers in different frames of reference. This can result in a change in the frequency and wavelength of the photon, as well as a change in its direction of travel.

## 3. Can angled photons violate the speed of light limit in special relativity?

No, angled photons cannot violate the speed of light limit in special relativity. According to special relativity, the speed of light is the same for all observers, regardless of their relative motion. Therefore, even though an angled photon may appear to be traveling faster or slower to different observers, its actual speed is always the same.

## 4. What are some real-world applications of angled photons in special relativity?

Angled photons are important in many areas of modern physics, including particle accelerators, high-energy physics experiments, and the study of the universe. They are also used in technologies such as lasers, fiber optics, and medical imaging devices.

## 5. How does the concept of angled photons in special relativity relate to the theory of relativity?

The concept of angled photons in special relativity is a direct result of the theory of relativity, which describes how the laws of physics are the same for all observers in inertial frames of reference. Special relativity allows us to understand how angled photons behave differently in different frames of reference, and how this is related to the concept of time and space being relative.

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