# Tag Info

17

Calculating the effect of acceleration in special relativity is straightforward, but I suspect the algebra is a bit much at high school level. See John Baez's article on the Relativistic Rocket for a summary, or see Chapter 6 of Gravitation by Misner, Thorne and Wheeler for a more detailed analysis. When you're first introduced to SR you tend to be told ...

6

Whether it's a black hole or some other more ordinary mass pulling on your rope isn't actually that interesting. Let's think about a cable unrolling above Earth to start with. What we have is a pulley with a rope hanging off one side. The weight of the rope exerts some force on the edge of the pulley, causing it to undergo angular acceleration (starts to ...

5

You can define the age of the universe roughly as the proper time for a hypothetical observer who is comoving with the galaxies and not too near a strongly gravitating object. This is imprecise because the galaxies are themselves moving around and the age would depend on exactly the worldline of the observer and how it moved to avoid heavy objects that ...

4

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$, $\vec{F}=\vec{F_1}+\vec{F_2}=m(\vec{a_1}+\vec{a_2})$ When using vectors symbol, its automatically takes care of their directions.

4

No the speed of light in vaccuum is an absolute constant $c$ = 299 792 458 m/s The way to add up relativistic speeds is: $u' = \frac{u-v}{1-\frac{uv}{c^2}}$ to account for the constancy of the speed of light You cannot simply add them up. Edit: This also applies to normal everyday speeds. The reason we don't use this formula is because the speeds we are ...

3

If you are the observer and you are observing a car move in front of you, then the clock in the car will appear to move slower than the clock in your reference frame and the length of the car will appear to be contracted to you in the direction of motion. All these effects get magnified as the velocity of the car approaches that of light. In the same way, ...

2

This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i$$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...

2

Galileo proposed that all inertial systems are equal. That means there is no absolute observer. He stated that if you sit in the belly of a ship which is moving on a calm sea, you cannot know or measure its velocity with respect to the ground of the ocean. Special relativity makes this statement a little different as it introduces the maximum speed $c$, but ...

2

The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$a=\frac{m_3-m_2}{m_2+m_3}g\tag{1}$$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...

2

Critical mass is actually more about 'the right number of nuclei in a specified space'. As we are talking about solid matter this equivalently translates to a given number of atoms (or molecules depending on your matter). And this furthermore translates to our everyday mass. But it's 'not about mass', it is just a practical way to specify the quantity. So ...

1

1 - It is false! If $E = mc^{2}$ is true only for an object that isn’t moving, the mass never changes (is a "Lorentz invariant"). 2 - Can you rephrase it, please? 3 - Energy and mass are not at all the same thing; an object’s energy can change when its motion changes, but its mass remains the same. 4 - In Special Relativity, time can be variable, its ...

1

The bomb doesn't "care" what its mass might look like to an observer in another frame. If you calculate critical mass you don't worry how big it might seem to observers located in billions of other possible frames of reference. Local frame of reference is the only one valid for making calculations concerning the occurence of local phenomena.

1

The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...

1

So, you are correct, that it is not quite as simple as that. Have you studied Special Relativity (SR) yet? One of the most fundamental ideas is the limiting speed of light. Nothing moves faster than c, and this means that velocities must add differently. Think of the reverse, if you were on the rocket-ship traveling at $0.5c$ and shot a laser-beam at $c$, ...

1

While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...

1

No. The standard metric for cosmology is given by: $$ds^{2} = - dt^{2} + a(t)^{2}\left(d^{3}{\vec x}^{2}\right)$$ where the term inside the parenthees represents the 3-metric of a homogenous three space. As you can see, there is no difficulty with evaluating the age of the universe: $$T = \int\sqrt{-g}\,\,x^{a}y^{a}z^{a}\epsilon_{abcd} = \int dt$$ ...

1

Basically, your idea that "...the combined action of both (i.e. that the Universe is expanding and an actual speed of removal of the galaxy)." is correct. The one due to expansion might be called "comoving velocity" while the one due to the "actual" motion of the galaxy through the expanding universe is usually called the "peculiar velocity". The observed ...

1

For any observer observing two events, $E_1$ and $E_2$, the Lorentz factor is: $$\gamma = \frac{\Delta t}{\Delta \tau} \tag{1}$$ where $\Delta t$ is the time interval measured by the observer, i.e. $t_2 - t_1$, and $\Delta \tau$ is the proper time difference calculated by the observer. To see this consider the example you give of our observer watching a ...

1

I think your confusion with the slingshot is this: when you move your hand in circles to keep the slingshot moving in a circle you need some force, so you feel the weight of the projectile. But you where taught that according to Newton, things want to keep going in a straight line and inertia is this tendency to "not want to move or change direction". So ...

1

Okay, let's lay some ground rules: An object that does not experience any force will fly in a straight line A force applied to an object will change its momentum toward the direction of the force. Now, the trick with circular motion is that both the direction of motion and the direction of the force change simultaneousely, such that the inward cetripetal ...

1

The direction of acceleration is the same direction for the propagation of inertia usually! This is where you are wrong. Direction of propagation is the direction in which the body is currently moving or rather, changing position. So it is the direction of infinitesimal displacement $d\vec x$ at that instant. Now what makes you think that direction of ...

1

Fictitious forces naturally arise in non-inertial (accelerating) reference frames and you have to be careful with them. In this example, it only leads to confusion. $F = ma$ tells us that when the (total) force is zero, the object will continue in a straight line. It's not the centripetal force, but the absence of a centripetal force that makes the object ...

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