Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A few questions about light

  1. Feb 26, 2014 #1
    Hi guys,

    Firstly apologies if this question has been asked before (or are stupid questions) or better explained elsewhere. I'm not studying physics nor pursuing a scientific career, just curious about the following questions.

    The questions I have are regarding the speed of light. Firstly if an object travelled at the speed of light or close to it, there would be two different speeds? One speed from a perspective outside the object and a second one from inside the object as the time travelled would be different?

    Secondly, if two objects were travelling apart at 50.01% speed of light, would I be correct in assuming the object would not be visible from spectating from one object to the other? e.g. if the Earth was travelling in one direction at 50.01% the speed of light and the Moon in the opposite direction at the same speed, the object would not be visible from Earth?

    Following on from that if two objects were travelling towards each other at 50.01% the speed of light would I be correct in assuming that object would also not be visible?

    Lastly, going back to two objects moving apart at 50.01% the speed of light, could one object be used as reference at looking at the other object travelling at 100.02% the speed of light?
     
  2. jcsd
  3. Feb 26, 2014 #2
    These are great questions! First off, it's important to know that speeds don't actually add in the way we normally treat them. This is because when we observe an object, we are actually observing light emitted by the object and thus, the fastest that information can travel is c. It's pretty clear, then that nothing can be observed to move faster than c. The other important fact to note before answering your questions is that there is no such thing as an absolute velocity. Velocity is always measured/stated with respect to a reference. So in your first example you ask what an observer on the first object would see if he looked at the other object. I'm assuming that in the setup when you said that they were both traveling at 0.5c in opposite directions, that this is measured in a reference frame that is between the two objects such that both objects appears to be traveling in opposite directions away from the center of reference frame. So now we have two reference frames - the original observation reference frame where the two objects are both moving away from the center at 0.5c and the reference frame that is on one of the objects. Without going into the derivation, i'm just going to tell you that when you want to change from one reference frame to another, the speeds change like this: v1-v2/(1-v1v2/c^2). This is due to the fact that all of the observations are happening by the transfer of light or other massless information between the two objects in question. So in this case, if you move from the first reference frame to the second reference frame (Which is moving at 0.5c with respect to the first frame -- as measured in the first frame), then you conclude that an observer on the first object would see the other object moving at 0.5c-(-0.5c)/(1-(0.5c*-0.5c)/c^2) == c/(1+0.25c^2/c^2) == 0.8c (where you will note that i made one of the 0.5c's negative since it's traveling in the opposite direction in the original reference frame). So an observer on one object sees the other object moving at 0.8c, even though an observer in the original reference frame sees both objects as moving as 0.5c in opposite directions.

    1) So the answer to your first question is yes, there are different speeds measured depending on how fast the observer is moving with respect to the objects.

    2) Since the observer never observes any massive objects moving faster than c, light can always reach the observer, so you will still see the moon from earth and vice versa. However, what WILL happen is that the light will redshift, so the moon will look redder the faster it is moving away from an observer on earth. This happens due to doppler shifting (very similar to the kind you hear with a passing siren) and also due to the expansion of space (stretching out the light waves as they travel through it -- assuming we're talking about big distances)

    3) They would still be visible to each other as before (you can calculate the apparent speed using the equation above if you want!), and they will appear blue shifted.

    4) Already answered that above. You would measure 0.8c, but you can certainly use it (or any other frame moving at constant velocity) as a reference!

    Hope this helps
    George
     
  4. Feb 26, 2014 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    http://en.wikipedia.org/wiki/Velocity-addition_formula
    Above describes how to add velocities of fast moving objects.

    First question: all speeds are relative.
    For your other questions, go to reference to see how the speeds add. They don't add linearly.
     
  5. Feb 26, 2014 #4

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    @ocset
    It would be useful for you to read something about Special Relativity - for instance, the Wiki pages. You will get a lot further, a lot quicker if you get yourself informed by some formal reading than by posting lists of questions about your particular take on these things. Q and A is a very inefficient way of learning until you have a reasonable basis on which to build - there will be too many random ideas that don't relate to reality if you start in a vacuum.

    Einstein's book on Special Relativity is available all over the place as a free e-book. Get it from the horse's mouth.
     
  6. Feb 27, 2014 #5
    As an example, using the addition formula [tex]v_{rel}=\frac{v_1+v_2}{1+\frac{v_1 v_2}{c^2}}[/tex], where [itex]v_1[/itex], and [itex]v_2[/itex] are the speeds of the two objects moving in opposite directions, c is the speed of light and [itex]v_{rel}[/itex] is the relative speed between the objects, and plugging in the values you suggested [itex]v_1 = 0.5001 c[/itex], and [itex]v_2 = 0.5001 c[/itex], you get [tex]v_{rel}=\frac{v_1+v_2}{1+\frac{v_1 v_2}{c^2}} = \frac{1.0002c}{1+0.5001^2} = 0.800096... c,[/tex] so the two objects see each other moving at about 80% of the speed of light.
     
  7. Feb 27, 2014 #6

    jtbell

    User Avatar

    Staff: Mentor

    Call those two objects A and B, and add a "stationary" object C. As people have already said, someone "riding" on either A or B would "see" the other object moving away at about 0.8c.

    People studying relativity would say this more precisely as "in the reference frame in which A is at rest, B has a velocity of about 0.8c" and similarly, switching A and B. Or more concisely, "in A's rest frame, B has a velocity of about 0.8c".

    Someone "riding" on C would "see" the distance between A and B increasing at the rate of 1.0002c. That is, in C's rest frame, the distance between A and B increases at the rate of 1.0002c. There's no problem with this in relativity, because it's not actually the velocity of either A or B in any reference frame.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook