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Spacetime interval formula - what's the d?

  1. Jan 27, 2005 #1
    I got this formula for calculating the distance between spacetime intervals at Wikipedia:

    ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2

    Now first I thought the d was supposed to be the Greek delta which my physics teacher uses to indicate a difference of the following variable, instead of just a simple value, but when looking at other sites, I saw they also used the d. So my question is, is this the same as the delta?

    Secondly, I was wondering if someone could explain this formula to me. I understand what it says (if the d is indeed what I thought it was - the delta), but not why this is the formula.
  2. jcsd
  3. Jan 27, 2005 #2


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    That's the differential/infinitesimal for of the square of relativistic interval,that's it...So [itex] \Delta [/itex] goes into [itex] d [/itex] each time very,very small (YET NONZERO) variations of the argument are considered...

  4. Jan 27, 2005 #3
    Thanks for replying, but...I'm afraid I have no idea what you just said.
  5. Jan 27, 2005 #4


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    Then wait until u'll learn (differential) calculus and everything will become amazingly simple...


    EDIT:Yeah,good idea,RAHITICAL DELTA... :rofl:
    Last edited: Jan 27, 2005
  6. Jan 27, 2005 #5


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    Dearly Missed

    As long as you haven't learnt calculus, just think of it as a tiny "delta".
  7. Jan 27, 2005 #6
    Thanks, arildno.

    Is it impossible to understand the answer to my second question without knowing this calculus?
  8. Jan 27, 2005 #7


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    In general, the 'd' means that the deltas have to be small for the formula to work correctly.

    Fortunately for you, in special relativity, when space-time is flat, the deltas don't even have to be small. So you can use the above formula "as-is", by thinking of the d's as deltas, as long as you are dealing only with special relativity. (This won't work if there is gravity involved, however, so some other more complicated formulas you see in notation simlilar to the above won't necessarily work with large detlas).

    To quantify the notion of "how small does small have to be?" in the more general case, you'd probably need to learn some calculus.
  9. Jan 27, 2005 #8
    d versus delta

    Hi Nana
    Your question deals with 'relativity' and as I understand it NO d's are allowed ---- the reason being that it implies changes due to forces .
    The normal equation has NO d's and just states the relations in 4 dimensional space , time and distance caused by the property of light.
    Under very mild accellerations ( ones which do not imply a lot of mass ( per E=m.c^2) ) then you can allow small variations --- so the 'd' represents small changes in all variables and as such means a very complex situation
    because no -one has stated HOW such variables are changing .
    In other words IFFF you introduce 'd' meaning samll changes it automatically implies there is a reason for the change -- But NOT stated .
    Special relativity does not really try to deal with this -- General relativity does -- but that is a whole different game -- if you are up to it . Ray
  10. Jan 27, 2005 #9


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    WHAT????????????Jesus,what are u talking about???What forces...???

    What??It has delta for finite variations and "d" for infinitesimal ones...What's "normal"??

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