Negative time from Lorentz transformation of event in S

[Orad]

See, I've just started taking this class, and evne though I've already read books by Martin Gardner and Brian Greene, I don't understand any of this stuff... well, not any. But my math seems to be wrong.

I have 2 frames, S and S', such that t=t'=0 and x=x'=0.
Event A occurs in frame S at tA=0.3 microseconds, xA = 150 m.
Frame S' moves at a velocity of +0.65c (where c is 3x10^8 m/s, by our convention)

I don't not understand what to do, but when I do the full lorentz transformation calculation, I end up with a negative time for t'A. Am I miss-interpretting the question or the answer? Why?

 

[George Jones]

I haven't performed the calculation, but why is it a problem if t'A is negative?

 

[Orad]

Well, if t is negative, that means the event is in the past, for the observer, right? Doesn't that mean he never observed it? I figure it doesn't mean the same thing as saying it's 3 meters in the negative x direction, as you can look back and see it, but if it's the past.

Although, now that you mention it, since this observer's (S') moving so quickly, he's already seen the event and moved on by the time the observer in S sees it.

So I miss-interpretted the answer...?

 

[George Jones]

I just did a quick run through the calculation. I could have made a mistake, but I, too, get t'_A to be negative.

There is nothing magical or mystical about this. For example, if I choose here and now to be the origin of my spacetime coordinates, what is the time coordinate of something that happened yesterday, like the landing of the space shuttle?

Also, be careful with way "observe" is used in relativity. It doesn't mean the same thing as "actually see happen."