2
$\begingroup$

I have questions in 2 scenarios regarding this.

  1. Lets say according to an observer P at rest with respect to the earth, me & my friend start at the same time & at the same height to free fall under gravity to the ground. Lets say P is standing near to the point where I would hit the ground. At the start height the horizontal distance (ie horizontal/parallel to the ground) between my friend & me is small so that the earth's gravitational field between us is uniform. My friend has a torch in his hand. Lets say the torch is only able to produce 1 photon when its switched on. At the midway of the journey, he switches on his torch. Does that observer P see the photon hit the ground first before my friend & me?

  2. Lets say according to an observer P at rest with respect to the earth, me & my friend start at the same time & at the same height to free fall under gravity to the ground. Lets say P is standing near to the point where I would hit the ground. Here at the start height the distance between my friend & me is very large over the surface of earth that the earth's gravitational field between us is not uniform. My friend has a torch in his hand. Lets say the torch is only able to produce 1 photon when its switched on.At the midway of the journey, he switches on his torch. Does that observer P see the photon hit the ground first before my friend & me? And would I (while falling) record the speed of that photon as c or something else?

$\endgroup$
  • $\begingroup$ @PM 2Ring- The torch is pointing downwards. $\endgroup$ – CuriousMind9 Oct 29 at 10:58
  • 1
    $\begingroup$ Of course the photon will hit the earth first, nobody can overtake a photon. And locally you always measure a photon to have exactly c, no matter how fast you are. $\endgroup$ – Yukterez Oct 29 at 11:16
  • $\begingroup$ @Yukterez- What about the 2nd question of 2nd scenario? $\endgroup$ – CuriousMind9 Oct 29 at 11:18
  • $\begingroup$ That depends on your distance of separation, if you are so far apart that the light travel time from observer A to observer B is longer than the free fall time you will see the effects of the photon hitting the ground after you hit the ground, and otherwise the other way round. Assuming that the photon which hit the ground reflects into your eye, otherwise you see nothing. $\endgroup$ – Yukterez Oct 29 at 11:21
  • $\begingroup$ 1. Why wouldn't the photon hit the ground first? 2. Light always has a local speed of c. OTOH, to measure the speed of the photon it has to pass through (& interact with) your measuring device; you can't measure a distant photon. And only 1 person can detect a single photon, all 3 observers can't see it. $\endgroup$ – PM 2Ring Oct 29 at 11:24
2
$\begingroup$

You do not need this complicated setup. You have a clear and straightforward question about the speed of light. I will answer that one. The other question " Does that observer P see the photon hit the ground first before my friend & me?", is hard to tell because of the setup, but basically, because seeing a photon hit the ground can only mean that there is a sensor that senses the photon hitting the ground, and then sending signals back about that fact to the observers, and since the observers in free fall are so slow compared to the speed of light, the one closest to the sensor will "see" the photon hit the ground first (meaning will receive info about it first).

Now you are asking about the speed of light, and your question is about local and non-local measurements.

The speed of light is always c when measured locally in vacuum. Now let's disregard the atmosphere, and let's talk about vacuum.

You are asking whether the nonlocal measurements of light can vary from c. Yes they can. If you measure the speed of light non-locally, you can measure speeds less and (yes contrary to popular belief) more then c.

Let's talk about the easier one first, the less then c speeds. This is called the Shapiro delay, and basically, when you measure (from Earth) the speed of light as it passes next to the Sun (where the gravitational potential is stronger), you will measure speeds less then c. This is because the clock next to the Sun ticks slower then your clock here on Earth. So you divide a distance with a time that is more (because more time passed on your clock here on Earth), so you get a speed less then c.

https://en.wikipedia.org/wiki/Shapiro_time_delay

Now let's talk about the harder one, speed more then c. Yes it is possible, but it is experimentally hard to do. You need to check the speed of light as it passes in empty space, and you are measuring it from a neutron star (where the gravitational potential is stronger). In this case, your clock will tick slower, you will divide the distance by a time that is less, so you will get a speed more then c.

$\endgroup$
1
$\begingroup$

There are several issues with the wording of your question, as follows:

1) You do not say which way the torch is pointing.

2) You talk about an observer being 'near your frame of reference'. I do not know what you mean by that. The observer is stationary on the ground, so their frame of reference is not yours.

3) When you ask whether one person 'sees' the photon hitting the ground, you must be talking figuratively, but one cannot know whether you mean 'record the time of arrival' or 'see the reflected light'.

4) One difference between your two scenarios is that you and your falling friend are in the same reference frame in the first, but you are in separate frames in the second, as your velocity will no longer be in the same direction as your friend's. The relativistic effects of time dilation etc will be entirely reciprocal, but your clocks will no longer be synchronised, so you have no way of knowing (in your ambiguous scenario) when your friend switched-on their torch.

That all said, the answer is as follows:

The arrival of the photon on the Earth's surface is a physical event- there can be no ambiguity about it.

Whether the observer's disagree about the the timing of the event depends on how their clocks are synchronised.

You will always record the speed of the photon as c, as will the other observers- that is the whole point of relativity.

$\endgroup$
  • $\begingroup$ @SolomonSlow Great correction. Thanks. Have edited my answer to stress the point. $\endgroup$ – Marco Ocram Oct 30 at 6:49
0
$\begingroup$

You need to do some math to get a precise answer. The velocity of the photon, as measured by the observer standing on the earth, depends on the gravitational potential at the location of the photon. This may cause the photon to have a velocity slightly greater than $c$ near the falling observer. This velocity is reduced to $c$ near the observer on the ground. For calculating the time needed for the photon to reach the ground, you need to integrate the G-potential over $x$ (the varying distance of the photon from the ground).

Moreover, EEP asserts that the falling guy(s) is a Minkowski observer, and thus, he can easily apply Lorentz transformation for the emitted photon. In his frame of reference, the photon always moves at $c$ whether or not the field is constant or varying. He also sees that the surface of the ground approaches him with a constant or varying acceleration. From his point of view, the time needed for the photon to reach the ground can easily be calculated by applying Lorentz transformation.

However, I recommend you to consider one of the observers, instead of standing on the ground, at rest on the tower where the two others (your friend and you) start to fall freely towards the ground in order for all observers to easily synchronize their clocks before the experiment is carried out.

$\endgroup$
  • 1
    $\begingroup$ This may cause the photon to have a velocity slightly greater than 𝑐 near the falling observer. – this can never happen. Due to the equivalence principles, all local observers see special relativity, which means that light always travels with $c$. $\endgroup$ – Prof. Legolasov Oct 29 at 12:53
  • $\begingroup$ The observer on the ground is not a local one as long as he tries to consider the behavior of a photon far away from him. He is local only for when he wants to investigate the behavior of the photon very close to him (in his vicinity) or along the same equipotential line as his. $\endgroup$ – Mohammad Javanshiry Oct 29 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.