Given some time like worldline, proper time is the time measured by a clock traveling along that worldline. It is given by ##\tau=\int_P \sqrt{1-v^2/c^2}dt##Zephaniah said:What is proper time? How can I solve proper time?
The questions you asked in your initial post were very general. Maybe it would help if you asked more specific questions. Is this your only discomfort with the basics of special relativity? Or are other aspects of special relativity limiting your ability to understand property time.Zephaniah said:Thank you for the explanations up there. For everyone's information, I am a college student and this topic "Proper Time" is my report. And I am really having a hard time how to understand this concept that is why I still don't have a concrete idea about this one.
haushofer said:If you take 2 events, you can define the elapsed time between them in different ways, and this will give you different answers in general. That's the idea of relativity. Does it mean that each observer is correct whatever they state of what they have observed?
My teacher told me that there is a computation for proper time and I as I browse in the internet the formulas are confusing. I think I need a step by step instruction to solve it.Chestermiller said:The questions you asked in your initial post were very general. Maybe it would help if you asked more specific questions. Is this your only discomfort with the basics of special relativity? Or are other aspects of special relativity limiting your ability to understand property time.
If the two observers are at rest relative to one another (and, thus, they are both physically present at both events) their clocks will both show the same elapsed proper time. If the two observers are in relative motion, then they can't both be physically present at both of the two events.Zephaniah said:Now, I understand that Proper time is what the observer's clock reads. Right? How about if there are 2 observers with their own clock at the same event but they come up with a conclusion that the event happen with a different time in there clock which one will I consider as proper time?
Dale gave the equation in post #3. Is this the equation you are finding complicated to solve? Which part are you finding complicated?Zephaniah said:My teacher told me that there is a computation for proper time and I as I browse in the internet the formulas are confusing. I think I need a step by step instruction to solve it.
Is it the same with inertial reference and relativity of simultaineity?Chestermiller said:If the two observers are at rest relative to one another (and, thus, they are both physically present at both events) their clocks will both show the same elapsed proper time. If the two observers are in relative motion, then they can't both be physically present at both of the two events.
Sorry, I have no idea what you are asking here. It seems to me your problems are much broader than just not understanding proper time.Zephaniah said:Is it the same with inertial reference and relativity of simultaineity?
Yes Sir. That formula is too complicated for me.Chestermiller said:Dale gave the equation in post #3. Is this the equation you are finding complicated to solve? Which part are you finding complicated?
In what way? Have you not had integral calculus?Zephaniah said:Yes Sir. That formula is too complicated for me.
Chestermiller said:Sorry, I have no idea what you are asking here. It seems to me your problems are much broader than just not understanding proper time.
Chestermiller said:In what way? Have you not had integral calculus?
Let's see your attempt to do this so far. What is your best shot at explaining this?Zephaniah said:My problem is how am I going to teach the concept of proper time to my classmates and how am I going to show some problem solving to them.
Chestermiller said:Let's see your attempt to do this so far. What is your best shot at explaining this?
If you confine attention to inertial frames of reference that are in relative motion with respect to one another with velocity v, then the equation simplifies to ##\Delta \tau=\sqrt{1-\left(\frac{v}{c}\right)^2}\Delta t##. Imagine that you have a single observer with a clock that is at rest in his frame of reference, and measures the time interval between the two events ##\Delta \tau## (he is physically present at both events). Imagine that this observer is moving with velocity v relative to a (stationary) group of observers strung out along the route from the first event to the second event, and the two observers physically present at the two events write down the times on their synchronized clocks at which the two events occur. They then get together and compare notes, and, when they do, they find that, according to their clocks, the time interval between the two events is ##\Delta t##. The equation above will tell you the relationship between ##\Delta \tau## and ##\Delta t## (which will not be the same).Zephaniah said:Yes Sir. I haven't learn integral calculus yet.
Only the observer who is at rest and personally observes the two events can be physically present at both of the events. The other observer you are referring to (in a different frame of reference) can be physically present at either of the events, but not both of them.Zephaniah said:Maybe I'll draw an illustration showing an observer inside the event and another observer outside the event. Then I'll ask them if who among the two know the proper time. Then the answer will be both have observed the event at a different time but both of them has the proper time in their own reference.
Chestermiller said:If you confine attention to inertial frames of reference that are in relative motion with respect to one another with velocity v, then the equation simplifies to ##\Delta \tau=\sqrt{1-\left(\frac{v}{c}\right)^2}\Delta t##. Imagine that you have a single observer with a clock that is at rest in his frame of reference, and measures the time interval between the two events ##\Delta \tau##. Imagine that this observer is moving with velocity v relative to a (stationary) group of observers strung out along the route from the first event to the second event, and the two observers physically present at the two events write down the times on their synchronized clocks at which the two events occur. They then get together and compare notes, and, when they do, they find that, according to their clocks, the time interval between the two events is ##\Delta t##. The equation above will tell you the relationship between ##\Delta \tau## and ##\Delta t## (which will not be the same).
Chestermiller said:Only the observer who is at rest and personally observes the two events can be physically present at both of the events. The other observer you are referring to (in a different frame of reference) can be physically present at either of the events, but not both of them.
OK. The single observer with the clock is traveling at 0.9c relative to the stationary group of observers (strung out along the route) with their clocks. The single observer who is physically present at the two events notes that the time interval between these two events is 1 hour. Now, for the group of observers strung out along the route with their clocks, when they get together and compare notes, what do they measure the time interval between the same two events to be?Zephaniah said:This one formula is much easier than before but I think I need to try this one first. May I ask if you have any problem solving for me to solve?
You can't just talk about one event. You need to talk about the time interval between two events. The actual times on their clocks don't matter. Only the time interval between the events matter. That's what we mean by proper time.Zephaniah said:What if they both see the event? I mean the other one is present at the event (lets call it observer 1) and the other one just saw the event (observer 2). Observer 1 says that the event happen at 7:00 am while observer 2 says that it occur at 7:05 am. Then where is the proper time?
Ibix said:Concepts you may find helpful:
- Interval ##\Delta s^2=c^2\Delta t^2-(\Delta x^2+\Delta y^2 +\Delta z^2)##
- Worldlines
- Block universe
- Events
- Minkowski diagram
The block universe is 4d spacetime. Events are points in space at a given time. Worldlines are lines joining events - your worldline joins all the events you passed through in your life. Interval is the generalisation of Pythagorean distance to spacetime. You may wish to look up the relationship between proper time and interval, and then think about what a worldline's proper time means.
Note also that "proper" in this context is being used in its original Latin sense of "one's own", rather like "property". Not in the modern English sense of "correct".
I can't solve this one. I don't know the value of change in time/delta t or whatever it is called.Chestermiller said:OK. The single observer with the clock is traveling at 0.9c relative to the stationary group of observers (strung out along the route) with their clocks. The single observer who is physically present at the two events notes that the time interval between these two events is 1 hour. Now, for the group of observers strung out along the route with their clocks, when they get together and compare notes, what do they measure the time interval between the same two events to be?