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Posts posted by NowThatWeKnow
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E=mc^2
in Relativity
[math]E=MC^2[/math]
[math]1 kg * C^2 = 8.98755179 * 10^{16} joules[/math]
What is [math]8.98755179 * 10^{16} joules[/math] used for?
Is this just a conversion from mass to energy or energy needed to accelerate it to a certain speed in a given time?
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The numbers you quoted from Baez' Relativistic Rocket article simply assume the spacecraft has a constant proper acceleration of 1g.
Exactly
If you read the article further you will see how much fuel is needed to accomplish this voyage, and that is assuming perfect conversion of mass to photons.
A starlight solar sail would be nice but it doesn't look like we should hold our breath waiting for that to be developed. We have a long time to go before this trip is made for several reasons.
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If the universe only exists "now" the twins would have to share the same "nows". If not they would pass into the past and future with respect to each other and the space twin could not return to the earths present.
That is the point being made by the twins paradox, the space twin does return to Earth's future according to the clock on the space ship. A 12 year 1G rocket ride for you would mean 100,000+ years will pass on Earth.
Merged post follows:
Consecutive posts mergedwhen you move away from where and when you made the map it will need corrections with the Lorentz factorTravelers would require a computer controlled time keeping system. There is no reason why you could not calculate Earth time while traveling. Just thinking out loud and not suggesting anything that would be practical.
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... as we know thrust comes from the blast and that comes from fuel . and of course fuel has a mass that mass connected with the spaceship .so with same thrust and a variable mass of fuel on that ship will make the spaceship continue at 1g until the fuel will be zero .
in that way it will be more logical .
I have to take my ticket in that journey .
I do not think changing mass from fuel consumption is considered in the numbers in my first post. You will notice it takes 2 years to reach .97c and then another 10 years to reach .99999..c . It seems that maybe the rocket accelerates slower and slower as the speed increases because of this so called relativistic mass.
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How high the the atmosphere, and again what type of thermometer? If you want something measure the temp of the atoms in the adjacent ensemble you will need to shield from more than the sun...
Personally, I would just like to get a ball park reading off any thermometer with a large scale when 150 miles up and shielded from the Sun's visible light. Would other then visible electomagnetic energy have a large impact on the temperature? When the temperature of the dark side of the moon is taken you don't have to stick the thermometer in the moon's surface to get a reading, do you?
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Off-topic: Why is Einstein's name so often misspelled?
Because all we remember is the "I before e, except after c" part and then not use spell check?
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each twin has a different perception of how "fast" time is passing. one twins "now" could be the other twins 5 minutes. how long does "now" last?
It could be instantaneous like a digital change. Using the theory of relativity you could calculate an exact time for every location in space and that could be used for a now time. Just like clocks for different time zones we could have clocks for different locations in the universe that would run at different speeds. They would be displaying a now time for different frames.
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No, temperature is a property of groups of particles.
So what would a thermometer read in earth orbit if shielded from the Sun? -233C like the dark side of the moon? It sounds like space may not have a temperature but anything it it could.
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"now" is when the event occurs, and is the same for both twins. They only dissagree as to its time.
Twin A's 2008 is twin B's 2009. So an event could happen in two different years yet happen at the same time or "now" as you say. You could demonstrate this after the trip was over. Each twin may disagree on the year of a recent event, but agree it happened 1 hour ago. However, from two different frames they will not agree. I can relate to your thought process but I am usually wrong.
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The point is a hypothetical "void" would not have a temperature of absolute zero, it would not have a temperature at all.
That being said, would you take a coat or t-shirt if you were going to visit this hypothetical "void"?
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What type of thermometer?...
Is there a thermometer that is not made of "particles"?
I was just having fun saying you could not measure the temperature of a true void because it would not be a true void if something was there to measure it. I think some are implying that this "true void" may be outside of our universe. I am not disagreeing with you.
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Since the twins continuously share the same "now’" they should be the same biological age, though their clocks would indicate vastly different ages.
That isn't the way the story goes. The atomic clock, windup clock and biological clock all agree.
http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm
You should read this site (insert your favorite word).
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You NEED particles to have a temperature, it's part of the definition.
So if you put a thermometer in a true void, it would not longer be a true void?
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For the reasons above we would not see any blue shifted galaxies outside our local group. Blue shifts only occur within a galaxy cluster where galaxies are orbiting their combined center of mass, and only when such galaxies are coming together for a collision.
What about colliding galaxy clusters? I would think they should show a blue shift galaxy from out side their own cluster.
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who said I want reach the moon to speed of light ?
I just want to accelerate it > 0 in noticeable way .
You mixed my merged reply to swansont and you together. I understand what you are saying.
Merged post follows:
Consecutive posts mergedNoThis isn't a Bell curve.
http://en.wikipedia.org/wiki/File:Lorentz_factor.svg
This is
http://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svg
I stand corrected. I was considering the first 40% of the bell curve and not the complete curve. Making up my own definitions is not good.
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thanks
Unfortunately there is not enough references in Turkish (in english-turkish dictionary)
Google found many online dictionaries. This on is not too bad.
http://www.yourdictionary.com/
Wikipedia has too much info sometimes.
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Gamma tends to infinity as you approach c. It's not Gaussian.
It isn't a bell curve?
http://en.wikipedia.org/wiki/Lorentz_factor#Values
You would maintain your speed if you shut the engines down, and if you restarted them you would begin to accelerate again.
So, would the acceleration of a constant Force rocket slow its acceleration as speed increased? And restarting the engines would be less then 1G?
Merged post follows:
Consecutive posts merged... if I start the machine do you think the moon start to move or not ?mass of the Moon * (the speed of light^2) = 6.61483812 × 10^39 joules. You will need a big rocket and please do not point it towards Earth.
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maybe it has to do with the one-way nature of time. even though it should look the same whether you think of it as the ship accelerating away from earth, or earth accelerating away from the ship,you got in the ship and did something so everything changed for you.
if we could calculate our "true" length contraction we would know our velocity relative to a universal frame, but due to our limited perception we can only measure the change in our length contraction
I think tomorrow we will see a reply that says "the acceleration broke the symmetry". I need to think about that I guess.
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next thing ya know yer sayin stuff like Υ
this is a guess but i'd say when you stopped accelerating you are in an inertial frame but your mass has increased so now you need more power for the same acceleration.
check out http://en.wikipedia.org/wiki/Momentum#Momentum_in_relativistic_mechanics
This math agrees with you I think:
p=y(rest mass)v ---- density=time dilation(rest mass)v
but
If you are at rest in your inertial frame, who is to say what your speed is. Why couldn't the length contraction you see come from the objects around you moving and not you. This keeps coming up to haunt me like the twin paradox. Whoever did the initial accelerating is moving but that makes no sense to me in a metric without an ether.
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but,in turkish , heat,meaning:ısı
heat units:
Calorie -Joule.(not degree)
I know,degree is not heat unit in english(or in other languages)
but,because of -a degree of heat-......I am confused
Degree - English
1. A unit division of a temperature scale.
2. A unit of latitude or longitude, equal to 1/360 of a great circle.
3. Relative intensity or amount, as of a quality or an attribute: a high degree of accuracy.
And it doesn't stop there.
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it would be nice if it were easier to tell whats in the lectures but they sure beat tv so i just watch the whole series(click related when you find one you like).
did you try http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
between that (really basic)page and the links under "see also" you can probably find any symbols you're looking for
The "Greek letters used in mathematics" link helped some. I will post a list I started in my next post hoping people will add to or make corrections. Thanks
Merged post follows:
Consecutive posts mergedGamma. It's the term that tells you the amount of time dilation or length contraction[math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]
What I think I am seeing (here and your other Force link) is Force required to accelerate is somewhat proportional to time dilation. Time dilation becomes very nonlinear (like a bell curve) as relativistic speeds are increased.
1a. Is that right?
1b. The acceleration of a constant Force rocket would slow its acceleration as speed increased?
2. At any point the engines were shut down the rocket would become an inertial frame and maintain speed relative to its surroundings (not counting space dust and gravity influence)?
3. Why couldn't you restart the engines and accelerate out of your inertia frame?
I started a list. If anyone sees any errors or would like to add to it please feel welcome.
p = Density
P = Momentum
Y = Lorentz factor
t = Proper time
T = Time in another frame
c = Speed of light
u = velocity as observed in the reference frame where time t is measured
v = velocity
V = Volume
F = Force
G = Gravitational constant
m = Mass
d = Distance
a = Acceleration
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Have you looked at the Wikipedia article?
In essence it is a collection of functions that describe the energy, momentum, pressure and sheer-stress of "stuff". In general relativity it acts as a source for gravity.
Yes I did look at wikipedia but after this thread it makes more sense to me. Thanks for your succinct answer. That is exactly what I was looking for when I started this thread.
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if you've got some time, and haven't noticed yet, you might search you tube for "math lectures" and physics lectures too. MIT is awesome for posting these lectures
That might work if I can find the right one. A glossary of the characters used in these equations would also be nice. I need to get up to speed on math if I am going to keep playing this game.
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Not at your beck and call, obviously.
Forces get tricky in relativity
http://en.wikipedia.org/wiki/Special_relativity#Force
If you look at the fourth equation in that section, you'll see that Force and acceleration no longer have a linear dependence.
2 hours and 4 minutes, not too bad. (you know I appreciate your help)
Some of that math I could probably actually do but my first problem is I do not know what each character represents. It wasn't in my "Relativity for dummies" book. I have figured out a few but is there a complete list some where?
Sorta like
c = speed of light
v = velocity
V = volume ??
funny looking Y = ??
Should I be looking at this problem like E=MC^2? The faster you go, the more energy it takes and that is why it isn't linear? I keep thinking that at any time you shut the engines off you would be at rest and inertia would keep you going. Starting your engines again would quickly get you to 1G acceleration. Sort of like starting over again. Arn't you either at rest or accelerating? Even when the math is done it only gives you an answer and not a good visual of what is happening or why. Thanks
Merged post follows:
Consecutive posts mergedit's better to explain it in a low motion physics . try to imagine your are in space .and there is a road ( resistant of road=0) and your are in a bicycle . you started moving your legs and spinning the pad .your bicycle will started moving and you will slowly begin spin the pad faster and faster so I have a five things .1- is there any limit of your spinning ?
2- did you reach your limit of spinning?
3- at point your are spinning the pad at that limit ? is your bicycle accelerating ?
4- now stop spinning the pad did your bicycle stop ( don't forget you are still in a space without resistant ) so there is only velocity
5- now spin the pad again do you feel your bicycle has any acceleration as you already reached your limit.
my point there is three states of your position .
less spinning
equal spinning
more spinning
so ( less or equal spinning will not accelerate your bicycle ) rather than ( more spinning will accelerate it for sure ).
spinning = force
((about the resistant things :if resistant exist it's only make you slower but if it's not exist it doesn't mean you will go faster you only will go in actual speed ))
You speak my kind of math.
Are you saying that less spinning and equal spinning would produce the same results (constant velocity)? And more spinning is needed for acceleration?
When you take all resistance away it would seem that more spinning would not require additional force, just a higher gear. I guess the part that confuses me is I see everything as at rest in space so force would be force. How else can you look at it in a metric without an ether? If you have any comments on the post above this one I would appreciate it. I relate to your words and appreciate it. I have my good areas but formal training in physics is not one of them, as you can tell.
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E=mc^2
in Relativity
Posted
Of course, we also have f=ma. Thanks folks, your only easy question this week.