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

B Special Relativity and Time Dilation

  1. Jul 27, 2017 #1
    I am trying to get this idea of Time dilation understood.

    If there exists only two objects in a Universe and one object is stationary and the other object is moving at 99 % the speed of light. Their clocks were both synchronized when both objects were stationary relative to each other. Then the second object instantaneously accelerated to 99 percent speed of light. I believe that the person located in the speeding object would see his clock moving normally and would not see that it is moving slower than the person in the object that is stationary. Let us say this speeding object person then moves at 99 % speed of light for 10 minutes and then instantaneously stops and checks his clock. Will the fast moving person's clock have lost some time and moved slower than the person in the stationary object?
  2. jcsd
  3. Jul 27, 2017 #2


    User Avatar
    Gold Member

    There is no such thing as "stationary" or "moving" except as relates to a reference object, so your scenario is confused. If two objects start out next to each other and stationary relative to each other and the one of them fires a rocket and moves away, they will BOTH see the other's clock as moving slow (as long as the motion is going on) and their own moving normally. Only if you bring them back together can you compare the differential aging. Differential aging is different from time dilation in that time dilation is just something that happens to an object you are observing, never to you but differential aging is a real difference in age. Google the Twin Paradox.
  4. Jul 27, 2017 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    To answer that question properly, one needs to consider how one comparses two clocks that are not at the same location. One can describe the details of this process, and perhaps at some point this will become necessary, but it may be premature to do this at the current time. The two points I'd like to make are these. The first point is that this process of comparing clocks that are not at the same place has some subtle aspects in special relativity that are not immediately obvious, enough that it merits a discussion of its own. The second point is that there is a simple modification of your question that eliminates the immediate need to be address this issue of how to compare clocks at different locations.

    The modification consists of having one clock stay at rest, and having the other clock accelerate (instantaneously, if you like, or over some finite time period , it makes no difference). This clock then it travels for a while, and it accelerates in the opposite direction, so that it heads back towards the first clock. When it reaches the original clock, it accelerates for a third time and stops. This whole procedure has a name, it's called "the Twin paradox". Whether or not more objects than these two clocks exist in the universe is basically irrelevant. If adding additional objects helps you to understand how it all works, by all means, add more objects to the universe to help figure out what is happening.

    Now we can compare the two clocks, which are side by side (so we have no need of answering the first question) When we do this, we find that the two clocks have different readings.

    As an aside, the name we give to the concept of time that excludes comparing clocks at different locations is called "proper time". You'll find a lot of threads on this topic, some of which are perhaps overly long.
  5. Jul 27, 2017 #4

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    If Object A is at rest and Object B is moving relative to it at a speed of ##0.99\ c## then it's equally valid to say that B is at rest and A is moving relative to it at a speed of ##0.99\ c##. The only difference is the observer's state of motion. If he's co-moving with A then A is at rest, if he's co-moving with B then B is at rest.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted