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Some Burning Questions about Time Dilation

  1. Aug 8, 2009 #1
    Firstly, just to confirm, the value for velocity in the equations to calculate time dilation is the speed at which two things move away or towards each other? So if some object was moving parallel to the other, they would measure no time dilation on each other?

    But if one was moving perpendicular to the other, would the first measure time dilation based on there speed or how fast they travel away from them?


    Secondly, in the time dilation equations what is the velocity based on? I ask because, if something is moving at a speed which display time dilation for someone else, that would also affect the velocity of that thing. If your time slows down 50% relative to the observer than your velocity would then be 50% slower. So how can the velocity of something be used to calculate time dilation when it is constantly slowing down the faster it goes.

    - Thanks in advance.
     
  2. jcsd
  3. Aug 8, 2009 #2

    Doc Al

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    What matters is the relative velocity of the objects. The time dilation seen by an observer in some frame depends on the speed of the moving clock relative to that frame. In your first example, there would be no relative velocity and thus no observed time dilation; in your second example, there would be a relative velocity.
    I don't understand what you mean. How does observing time dilation (or anything else) affect the speed of the thing being observed?
     
  4. Aug 8, 2009 #3

    russ_watters

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    He's saying that time dilation affects the calculation of velocity. Which is true...but....

    There was once a tv commercial for spaghetti sauce: "It's in there!"
     
  5. Aug 8, 2009 #4
    the velocity in this equation is the velocity of an object as measured in the reference frame. this will tell you the time dilation experienced on the object, relative to the time experienced in that reference frame.

    if some objects where moving parallel to each other, then it is the case where in the reference from of one of the objects, the other object is at rest, in which case the velocity in the equation of time dilation is 0 so they experience no time dilation relative to each other.


    so, let's say in our reference frame, an object is moving to the east and another object is moving north... both at the same speed.

    in the reference frame of the object moving to the east, the other object is observed moving "north-west" at a speed a factor of sqrt(2) the speed observed in "our frame"

    the frame in which one is taking measurements and observations is a very important concept in SR and you must be thinking about it at all times. turns out most of these concepts like velocity, time, mass, etc. do not have definite values, and are only rigorously defined to the specific inertial frame at which they were measured.


    a general rule of thumb... everything in Special Relativity makes sense... we just often don't grow up thinking in these terms and so we find ways to get confused like this all the time...

    it is a great question because it is very important in solving problems to know exactly what each variable is... because as mentioned before, all of these things only have defined values in specific inertial frames.

    let's look at the equation:

    [tex]\Delta t' = \frac{\Delta t}{\sqrt{1-v^2 / c^2}}[/tex]

    in this case we are finding the time [tex]\Delta t'[/tex] that something in the "prime" reference frame (denoted by the ') would experience for an event that took a time [tex] \Delta t[/tex] in the other "normal" reference frame (denoted by a lack of ').

    So in this case, the v is the velocity of the prime reference frame as measured in the "normal" reference frame. the [tex] \Delta t[/tex] is the elapsed time for the event to transpire, as measured/observed in the "normal" reference frame.

    So even though time is kind of different in the "prime" reference frame, it does not matter because you use the velocity measured in the "normal" reference frame.
     
  6. Aug 8, 2009 #5

    Doc Al

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    Not exactly. (What if the speeds were 0.8c?) To find the relative velocity, you need to use the relativistic velocity transforms.
     
  7. Aug 8, 2009 #6
    haha, precisely - perhaps i should have taken my own advice when writing that.

    thanks for the correction :)
     
  8. Aug 8, 2009 #7

    A.T.

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    On coordinate velocity, which is based on the observers clock.
    Not coordinate velocity, but proper velocity, which is based on the dilated moving clock:
    http://en.wikipedia.org/wiki/Proper_velocity
    No, less time means higher velocity, when distance is the same. With 50% time dilation proper velocity is twice coordinate velocity.
     
    Last edited: Aug 9, 2009
  9. Aug 8, 2009 #8
    I thought this was very likely the case, would you mind pointing this portion out to me?


    Thank you! I finally have something to call these situations.

    1. I don't understand the connection your trying to make. I'd really appreciated an explanation as to why time dilated velocity doesn't effect the equations. Regardless of it being faster speed from slow down, or slower speed, it would seem to make some impact.

    2. That seems counter-intuitive... If time moves slower objects aren't going to go faster, they will slow down... But as you said, the proper velocity is faster, but that's not what I'm talking about. I'm talking about how it appears to the observer.

    But there's the point, what velocity is used to calculate time dilation? It would have to be the proper velocity, no?
     
  10. Aug 9, 2009 #9

    Doc Al

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    No, just plain old coordinate velocity. (At least if you use the standard "time dilation" formula of Δt = γΔt'; the speed used in calculating γ is the coordinate velocity.)
     
  11. Aug 9, 2009 #10

    A.T.

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    What is "dilated velocity"? The coordinate velocity is measured by a clock at rest to the observer, which is not dilated.
    Time doesn't move, it is observed to pass slower in the moving frame. The processes in the moving frame are observed as slowed down, if you use the fast clock of the observer to measure them. The velocity of the moving frame is higher if you use the slow clock to measure it. If your intuition tells you that dv/dt get smaller when dt gets smaller, then you intuition is wrong.
    No, coordinate velocity. I edited my first sentence in post #7 to make it more clear.
     
    Last edited: Aug 9, 2009
  12. Aug 9, 2009 #11
    (Numbers based on paragraph order)

    1. But when the observer measures the speed of the 'observee' it would appear to be less as it travel faster since time slows down for the observee...

    2. Doesn't the observer measure the dt of the observee to increase, therefore causing dv/dt to get smaller?

    3. It seems I don't quite understand what Coordinate Velocity is exactly. As I read this I have thought it was the velocity as seen by the observer because a few post up someone said something along the lines of: "proper velocity doesn't change, Coordinate Velocity doubles if the observee is time dilated by 50%."

    I'm going to study some of these terms more closely, I would appreciate some more explanation. Thanks.
     
  13. Aug 9, 2009 #12

    A.T.

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    No, the observer measures the moving clock to run slower, than his own.
    No, read it again.
    There is no more to explain. You need to properly read what has been explained to you.
     
  14. Aug 9, 2009 #13
    https://www.physicsforums.com/tags.php?tag=relativity+101 [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. Aug 9, 2009 #14
    Slower time equals more time...

    Oh right sorry, I didn't read that correctly. :)

    I'm still trying to get a handle on this. But I'd appreciate more input on this matter.
     
  16. Aug 10, 2009 #15
    Hi,
    I thought about this also, its realy confusing, because as the time slows down every thing also should appeare to us slow, even the heart rate, the movements and also the velocity, but what if we take space contraction also into account as the time appears to us slowed down the space also appears to us contracted so the net result of the velocity will be the same.
     
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